Oxidation state of $M n$ in $M n O_{4}^{-}$ is:

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Solution

Let $x$ be the oxidation state of $M n$ in $M n O_{4}^{-}$ .
Since the overall charge on the complex is $- 1$ , the sum of oxidation states of all elements in it should be equal to $- 1$ .
Therefore, $x + 4 (- 2) = - 1$
or, $x = + 7$

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